Determine the equation of the tangent line to the curve $\displaystyle y = \frac{e^x}{x}$ at the point $(1,e)$
Solving for the slope
$
\begin{equation}
\begin{aligned}
y' =& \frac{d}{dx} \left( \frac{e^x}{x} \right)
\\
\\
y' =& \frac{\displaystyle x \frac{d}{dx} (e^x) - (e^x) \frac{d}{dx} (x) }{x^2}
\\
\\
y' =& \frac{xe^x - e^x}{x^2}
\\
\\
y' =& \frac{e^x (x - 1)}{x^2}
\\
\\
y' =& \frac{e^1 (1 - 1)}{(1)^2}
\\
\\
y' =& \frac{e^1 (0)}{1}
\\
\\
y' =& \frac{0}{1}
\\
\\
y' =& 0
\end{aligned}
\end{equation}
$
Using Point Slope Form
$
\begin{equation}
\begin{aligned}
y - y_1 =& m(x - x_1)
\\
\\
y - e =& 0 (x - 1)
\\
\\
y - e =& 0
\\
\\
y =& e
\end{aligned}
\end{equation}
$
Equation of the tangent line at $(1, e)$
Sunday, October 25, 2015
Single Variable Calculus, Chapter 7, 7.3-2, Section 7.3-2, Problem 50
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